Question

Previously,

4%

of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than

4%

today. She randomly selects

125125

pregnant mothers and finds that

4

of them smoked 21 or more cigarettes during pregnancy. Test the researcher's statement at the

alpha equals 0.05α=0.05

level of significance.

What are the null and alternative hypotheses?

Upper H 0H0:

muμ

equals=

nothing versus

Upper H 1H1:

muμ

less than<

nothing

(Type integers or decimals. Do not round.)

Because np 0 left parenthesis 1 minus p 0 right parenthesisnp01−p0equals=nothing

▼

10, the normal model

▼

may not

may

be used to approximate the P-value.

(Round to one decimal place as needed.)

Find the P-value.

P-valueequals=nothing

(Round to three decimal places as needed.)

Is there sufficient evidence to support the obstetrician's statement?

A.

NoNo,

rejectreject

the null hypothesis because the P-value is

lessless

than

alphaα.

There

is notis not

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

4%.

B.

YesYes,

do not rejectdo not reject

the null hypothesis because the P-value is

greatergreater

than

alphaα.

There

isis

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

4%.

C.

NoNo,

do not rejectdo not reject

the null hypothesis because the P-value is

greatergreater

than

alphaα.

There

is notis not

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

4%.

D.

YesYes,

rejectreject

the null hypothesis because the P-value is

lessless

than

alphaα.

There

isis

44%.

Answer #1

We are testing,

H0: p= 0.04 vs H1: p<0.04

Now, n= 125, np= 5, n(1-p) = 120

Since np<10, the normal may not be used to approximate the p-value.

However assuming it's Correct:

z statistic: -p/√p(1-p)/n

Now, = 4/125= 0.032

So test statistic: (0.032-0.04)/√0.04*0.96/125= -4.56

p-value: P(z<-4.56) <0.00001

Since the p-value <significance level of 0.05, we have sufficient evidence to Reject H0 at the 5% level of significance and conclude that yes p<0.04

So option D is correct.

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smoked 21 or more cigarettes during pregnancy. Test the
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What are the null and alternative hypotheses?
H0:p =0.05 versus H1: p <0.05
(Type integers or decimals. Do not round.)...

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